Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Algebra worksheet — GCSE Higher
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- 1.Which expression means 'triple c, then add double d'?
- 2.Three vertices of a rectangle are (−4, −1), (2, −1) and (2, 3). The sides of the rectangle are parallel to the axes. Write down the coordinates of the fourth vertex.
- 3.Work out the gradient of the straight line with equation 2x + y = 8.
- 4.By completing the square, find the turning point of the curve y = 3x² + 12x + 7.y = 3x² + 12x + 7
- 5.The equation x² + 2x − 5 = 0 can be solved using the iterative formula xₙ₊₁ = 5/(xₙ + 2). The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₂ correct to 2 decimal places.
- 6.f(x) = x³ − 3x − 5. Given that f(2.2) = −0.952 and f(2.3) = 0.267, work out what this shows about the equation x³ − 3x − 5 = 0.y = x
- 7.A geometric sequence has first term 4 and common ratio √2. Work out the sum of the first three terms of the sequence, giving your answer in the form a + b√2.
- 8.An exponential model has equation y = A × bˣ, where b > 1. Its graph passes through the points (1, 135) and (2, 405). Work out the value of A.
- 9.The perimeter, P, of a rectangle with length l and width w is calculated using P = 2l + 2w. What kind of mathematical statement is P = 2l + 2w?
- 10.A student is asked whether the lines y = (2/5)x − 1 and 5y = −2x + 15 are perpendicular. The student answers: "Yes, because when I rearrange the second equation I get y = −(2/5)x + 3, and 2/5 times −2/5 is negative." Which of these is a correct comment on the student's reasoning?y = -2x + 15
- 11.Expand and simplify (x + 1)(x + 2)(x + 3).
- 12.A region R satisfies y ≤ x + 1, x + y ≤ 5 and y ≥ 0. Which of these correctly describes R in words?
- 13.A student attempts to prove that the sum of any three consecutive integers is a multiple of 3. Line 1: Let the three consecutive integers be n, n + 1 and n + 2. Line 2: Their sum is n + (n + 1) + (n + 2) = 3n + 2. Line 3: 3n + 2 leaves a remainder of 2 when divided by 3, so it is not a multiple of 3. Line 4: So the sum of three consecutive integers is not always a multiple of 3. Which line contains the FIRST error?
- 14.Work out the value of n² − n when n = −3
- 15.The formula for converting a temperature in Fahrenheit, F, to Celsius, C, is C = 5(F − 32) ÷ 9. Work out C, to 1 decimal place, when F = 98.6.
Answer key
- (c) 3c + 2d — Method: 'triple c' is 3c, 'double d' is 2d, and 'add' joins the two separate terms with a plus sign. Working: 3c + 2d. Answer: 3c + 2d. 2c + 3d comes from swapping which letter gets tripled and which gets doubled. 6cd comes from multiplying the two terms together instead of adding them, and also multiplying the coefficients (3 × 2 = 6). 5(c + d) comes from adding the coefficients (3 + 2 = 5) and applying that single number to both letters together, as if c and d always came as a pair.
- (b) (−4, 3) — Method: in a rectangle whose sides are parallel to the axes only two different x-coordinates and two different y-coordinates appear, and each of them is shared by a pair of vertices, so the missing vertex takes the x-coordinate and the y-coordinate that so far appear only once. Working: the x-coordinates given are −4, 2 and 2, so 2 is already used twice and −4 is used once; the y-coordinates given are −1, −1 and 3, so −1 is already used twice and 3 is used once; the fourth vertex therefore has x = −4 and y = 3. Answer: (−4, 3). The distractors: (3, −4) comes from picking the two unpaired coordinates correctly and then writing them in the wrong order; (−4, −5) comes from matching the 4-unit vertical side but measuring it downwards from (−4, −1) instead of upwards; (8, 3) comes from carrying on round the shape with the horizontal step used earlier, adding 6 to the x-coordinate of (2, 3) instead of closing the rectangle.
- (d) −2 — Method: a gradient can only be read off an equation that is written in the form y = mx + c, so y has to be made the subject first. Working: subtracting 2x from both sides of 2x + y = 8 gives y = −2x + 8, and comparing that with y = mx + c gives m = −2. Answer: −2. The distractors: 2 comes from reading the coefficient of x straight off the equation as it is printed, without rearranging, so the change of sign is missed; 8 comes from reading the constant as the gradient, confusing m with c; −1/2 comes from rearranging correctly and then writing the gradient upside down, as the change in x over the change in y.
- (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −5, confirming the minimum value −5 at x = −2: turning point x = −2, y = −5. Halving b instead of halving b/a — using −4 as the shift instead of −2 — lands on turning point x = −4, y = −41, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = 2, y = −5 — wrong, because (x + 2)² is zero at x = −2, not x = 2. Computing 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
- (d) It has a solution between x = 2.2 and x = 2.3 — Since f(2.2) is negative and f(2.3) is positive, the graph of f crosses the x-axis somewhere between x = 2.2 and x = 2.3, so the equation has a solution in that interval. Choosing 'between x = −2.2 and x = −2.3' confuses the negative f-VALUE at 2.2 with a negative x-value. Choosing 'no solution' misreads a change of sign as meaning the opposite of what it shows. Choosing 'exactly two solutions' assumes a single change of sign must give two roots, which is not what the rule guarantees.
- (a) 12 + 4√2 — a₁ = 4, a₂ = 4√2, and a₃ = a₁ × r² = 4 × (√2)² = 4 × 2 = 8. The sum of the first three terms is 4 + 4√2 + 8 = 12 + 4√2. 8 + 4√2 comes from leaving out a₁ and adding only a₂ + a₃ = 4√2 + 8. 36 + 4√2 comes from squaring the whole second term instead of applying the ratio to the first term: (4√2)² = 32 used as a₃, giving 4 + 4√2 + 32 = 36 + 4√2. 4 + 12√2 comes from using r³ instead of r² for the third term: 4 × (√2)³ = 4 × 2√2 = 8√2, giving 4 + 4√2 + 8√2 = 4 + 12√2.
- (a) 45 — The ratio between the two given points, one power of x apart, gives b: 405 ÷ 135 = 3, so b = 3. Substituting back, at x = 1, y = A × b, so 135 = A × 3, giving A = 45. Giving the common ratio b itself instead of A confuses which unknown was asked for and produces 3 — wrong, because the question asks for A, not b. Giving 135 instead treats the first given point as the y-intercept and reads A off it directly — wrong, because that point is at x = 1, not x = 0, so 135 is A × b, not A. Assuming A equals b⁰ = 1 by itself, rather than substituting a known point to solve for A, gives 1 — wrong, because b⁰ is always 1 regardless of A; A must be found using an actual (x, y) pair from the graph.
- (c) a formula — A statement that shows how to calculate one quantity from others, using an equals sign, is a formula. P = 2l + 2w tells us how to work out the perimeter, P, from the length and width, so it is a formula. Distractor origins: 'an equation' confuses a formula linking several letters with an equation solved for one unknown value; 'an identity' confuses a formula with a statement that is true for every value of a single variable; 'an expression' forgets that an expression has no equals sign at all.
- (c) Wrong — the product is −4/25, not −1. — The student's rearrangement is correct: 5y = −2x + 15 gives y = −(2/5)x + 3, gradient −2/5. But the test for perpendicularity is that the product of the two gradients equals exactly −1, not merely that it is negative. Here 2/5 × (−2/5) = −4/25, which is not −1, so the lines are NOT perpendicular. Distractor routes: "a negative product always means perpendicular" states a rule that does not exist — many pairs of lines have a negative gradient product without being perpendicular, as this pair shows. "The rearrangement is incorrect" wrongly blames a correct step; 5y = −2x + 15 does rearrange to y = −(2/5)x + 3. "2/5 and −2/5 are negatives of each other" notices a true but irrelevant fact — being negatives of each other is not the perpendicularity condition; an exact product of −1 is.
- (d) x³ + 6x² + 11x + 6 — (x + 1)(x + 2) = x² + 3x + 2. Multiplying by (x + 3): (x² + 3x + 2)(x + 3) = x³ + 3x² + 2x + 3x² + 9x + 6, which simplifies to x³ + 6x² + 11x + 6. Choosing x³ + 6x² + 6x + 6 has the right x² and constant terms but adds 1 + 2 + 3 = 6 for the x-coefficient instead of the correct sum of pairwise products 1×2 + 1×3 + 2×3 = 11. Choosing x³ + 5x² + 11x + 6 sums only two of the three constants (2 + 3 = 5) for the x² coefficient, leaving out the 1. Choosing x³ + 6x² + 11x + 5 adds the last two constants (2 + 3 = 5) instead of multiplying all three (1 × 2 × 3 = 6) for the constant term.
- (d) Below y = x + 1, below x + y = 5, above y = 0 — For y ≤ x + 1, R lies on or below the line y = x + 1. For x + y ≤ 5 (that is, y ≤ 5 − x), R lies on or below that line too. For y ≥ 0, R lies on or above the x-axis. Combining all three: R is below y = x + 1, below x + y = 5, and above y = 0. Distractor routes: "Above y = x + 1" flips the first inequality, describing the wrong side of that line. "Above x + y = 5" flips the second inequality, describing the wrong side of that line. "Below y = 0" flips the third inequality, describing the wrong side of the x-axis.
- (d) line 2 — Line 1 correctly represents three consecutive integers using n. Line 2 adds them: n + (n + 1) + (n + 2). Collecting terms: the n-terms give 3n, and the constants give 1 + 2 = 3, so the correct sum is 3n + 3, not 3n + 2 as Line 2 states — this is the first error, an arithmetic slip in collecting the constant terms. Lines 3 and 4 both follow correctly from Line 2's incorrect result, but that result itself is wrong: the true sum, 3n + 3 = 3(n + 1), is a multiple of 3 for every whole number n. Check the working of each line against what came before it, in order, rather than judging whether the final conclusion feels right — an error that flips the conclusion can sit several lines before the line that states it.
- (d) 12 — Method: substitute the value into both terms, remembering that subtracting a negative number has the same effect as adding the matching positive number. Working: n² = (−3) × (−3) = 9, and subtracting n means subtracting −3, which adds 3, so the calculation is 9 + 3 = 12. Answer: 12. The distractors: 6 comes from subtracting 3 rather than subtracting −3, giving 9 − 3; −6 comes from squaring −3 as −9 while still adding the 3, giving −9 + 3; −3 comes from reading n² as 2n, giving 2 × (−3) = −6 and then −6 + 3.
- (d) 37.0 — C = 5(98.6 − 32) ÷ 9 = 5 × 66.6 ÷ 9 = 333 ÷ 9 = 37.0. A candidate who forgets to subtract 32 first gets 5 × 98.6 ÷ 9 = 54.8 (1 d.p.). A candidate who forgets the 5 ÷ 9 factor entirely and just works out F − 32 gets 66.6. A candidate who multiplies by 9 ÷ 5 instead of 5 ÷ 9 gets 66.6 × 9 ÷ 5 = 119.9 (1 d.p.).
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